Euler's Substitutions for the Integral of a Particular Function

Euler's substitutions transform an integral of the form , where is a rational function of two arguments, into an integral of a rational function in the variable . Euler's second and third substitutions select a point on the curve according to a method dependent on the parameter values and make the parameter in the parametrized family of lines through that point. Euler's first substitution, used in the case where the curve is a hyperbola, lets be the intercept of a line parallel to one of the asymptotes of the curve. This Demonstration shows these curves and lines.
In symbolic calculations, the Demonstration shows:
1. If , the substitution can be . We only consider the case .
2. If , where and are real numbers, the substitution is .
3. If , the substitution can be . We only consider the case .
In all three cases, a linear equation for in terms of is obtained. So , , and are rational expressions in .


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Consider the curve (1) and a point on it. The straight line (2) through intersects the curve in another point . Eliminating from (1) and (2) gives
From that, and since
(3) becomes
which simplifies to
So is a rational function of , is a rational function of , and because of (2), is a rational function of .
So the relation
defines the substitution that rationalizes the integral.
Suppose that the trinomial has a real root . Then we get Euler's second substitution taking ,
If , then the curve intersects the axis at , which must be the point . This is Euler's third substitution
In the case of Euler's first substitution, the point is at infinity, , so the curve is a hyperbola. An asymptote is . We are looking for the intersection of the curve by straight lines that are parallel to the asymptote. The intersection of such a line gives a point , which is rational in terms of . This gives Euler's first substitution
[1] G. M. Fihtenholjc, Lectures in Differential and Integral Calculus (in Russian), Vol. 2, Moscow: Nauka, 1966 pp. 56–60.
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